Đáp án:
\(\begin{array}{l}
*)\\
\left[ \begin{array}{l}
x = k\pi \\
x = \dfrac{\pi }{4} + k\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
*)\\
\left[ \begin{array}{l}
x = 70^\circ + k.120^\circ \\
x = - 210^\circ + k.360^\circ
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
*)\\
\sin x.\left( {\sin 2x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\sin 2x - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\sin 2x = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
2x = \dfrac{\pi }{2} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \dfrac{\pi }{4} + k\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
*)\\
\sin \left( {x - 120^\circ } \right) - \cos 2x = 0\\
\Leftrightarrow \cos 2x = \sin \left( {x - 120^\circ } \right)\\
\Leftrightarrow \cos 2x = \cos \left[ {90^\circ - \left( {x - 120^\circ } \right)} \right]\\
\Leftrightarrow \cos 2x = \cos \left( {210^\circ - x} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
2x = 210^\circ - x + k.360^\circ \\
2x = x - 210^\circ + k.360^\circ
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
3x = 210^\circ + k.360^\circ \\
x = - 210^\circ + k.360^\circ
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 70^\circ + k.120^\circ \\
x = - 210^\circ + k.360^\circ
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
